# Set of points where stalk is integral domain is open

I struggled to find a solution for the exercise 4.9 in the second chapter of Liu's book Algebraic Geometry and Arithmetic Curves.

The first part is to show the set of points $$x\in X$$ such that $$\mathcal{O}_{X,x}$$ is reduced is an open subset when $$X$$ is a locally Noetherian scheme. This is not a problem because since $$X$$ is locally Noetherian we can assume to work with an affine Noetherian scheme $$\text{Spec}A$$, where $$A$$ is a Noetherian ring, and $$\{\mathfrak{p}\in\text{Spec}A | A_{\mathfrak{p}} \text{is reduced}\}=\{\mathfrak{p}\in\text{Spec}A | (N_A)_{\mathfrak{p}}=0\}=\text{Spec}A-\text{supp}(N_A)=\text{Spec}A-V(Ann(N_A))$$ and this is open.

The second part is the same thing with integral domains, show the set of points $$x\in X$$ such that $$\mathcal{O}_{X,x}$$ is an integral domain is an open subset when $$X$$ is a locally Noetherian scheme. This is where my problems start. As above we can assume to work with an affine and Noetherian scheme $$\text{Spec}A$$ with $$A$$ Noetherian ring. I think I have to find a characterization of the prime ideals where the localization is an integral domain, something like the reduced case.

After hours of attemps, my ideas are:

• if $$\frac{a}{s}\frac{b}{t}=0\in A_{\mathfrak{p}}\Leftrightarrow$$ exists $$u\in A-\mathfrak{p}$$ such that $$abu=0$$ in $$A$$ and so $$a,b$$ must be zero-divisors in $$A$$.
• if $$a,b\ne0$$ in $$A$$ and $$ab=0$$, in $$A_{\mathfrak{p}}$$ at least one of $$a,b$$ must be zero in $$A_{\mathfrak{p}}$$, otherwise $$\frac{a}{1}\frac{b}{1}=0$$, and so $$\mathfrak{p}\notin \text{supp}(a)$$ or $$\mathfrak{p}\notin\text{supp}(b)$$.

If we take $$D$$ the set of the zero-divisors of $$A$$ without zero, $$M_{(a,b)}=\text{supp}(a)^c\cup \text{supp}(b)^c=(\text{supp}(a)\cap \text{supp}(b))^c, \quad \forall a,b\in D$$ then each $$M_{(a,b)}=\text{Spec}A-V(Ann(a)+Ann(b))$$ is open in $$\text{Spec} A$$.

This is my claim: $$U=\bigcap_{a,b\in D}M_{(a,b)}$$ is exactly the set of prime ideals such that the localization is an integral domain. If $$\mathfrak{p}$$ is such that $$A_{\mathfrak{p}}$$ is an integral domain, then $$\forall a,b\in D$$ necessarily $$\mathfrak{p}\in(\text{supp}(a)\cap \text{supp}(b))^c=M_{(a,b)}$$. If $$\mathfrak{p}$$ is such that $$A_{\mathfrak{p}}$$ is not an integral domain, then there exist $$\frac{a}{s},\frac{b}{t}\in A_{\mathfrak{p}}$$ such that $$\frac{a}{s}\frac{b}{t}=0$$ and hence $$a,b$$ are zero-divisors of $$A$$ and $$\mathfrak{p}\in\text{supp}(a)\cap \text{supp}(b)=M_{(a,b)}^c$$ and $$\mathfrak{p}\notin U$$.

The problem of this construction is that the set $$U$$ is not open and I can't conclude.

Is this the right way to follow? Could someone give me some hints or advices? Thanks in advance to those who can answer me.

• Hint: show that if $A$ is noetherian, and $\mathfrak{p}$ is a prime ideal, there is some $f \notin \mathfrak{p}$ such that $A_f \rightarrow A_{\mathfrak{p}}$ is injective. Mar 28 at 9:50
• I don't understand the meaning of your title. Mar 28 at 10:09
• @Mindlack of course this implies $A_{\mathfrak{p}}\in D(f)$ but is this principal open subset made of primes whose localization is an integral domain? Mar 28 at 10:15
• @JeanMarie I'm sorry I don't know how to improve it... For sure it is clear reading the content Mar 28 at 10:24
• @Mindlack Why is $f$ so special? $f$ can be also $1$... Mar 28 at 10:24

As you said, let $$A$$ be a Noetherian ring. Remember that a ring is an integral domain if and only if it is reduced and has only one minimal prime. On the other hand, minimal primes of $$A_{\mathfrak p}$$ are minimal primes of $$A$$ contained in $$\mathfrak p$$.

Since $$A$$ is Noetherian, there are only finitely many minimal primes $$\mathfrak p_1, \ldots , \mathfrak p_n$$ in $$A$$, so $$A_{\mathfrak p}$$ has at least two minimal primes if and only if there are $$i \neq j$$ such that $$\mathfrak p_i + \mathfrak p_j \subset \mathfrak p$$, or equivalently, if $$\mathfrak p \in V(\mathfrak p_i + \mathfrak p_j)$$. Thus, $$\{\mathfrak p \mid A_\mathfrak p \text{ has one minimal prime}\} = \operatorname{Spec}A \smallsetminus \bigcup_{i \neq j} V(\mathfrak p_i + \mathfrak p_j)$$ is open, and by the first comment of my post, $$\{\mathfrak p \mid A_\mathfrak p \text{ is an integral domain}\} = \{\mathfrak p \mid A_\mathfrak p \text{ is reduced}\} \cap \{\mathfrak p \mid A_\mathfrak p \text{ has one minimal prime}\}$$ is also open.

• Thank you! Really simple solution! Mar 28 at 16:28

I’m turning my comment into an answer. Let $$A$$ be a Noetherian ring, and $$\mathfrak{p}$$ be a prime ideal such that $$A_{\mathfrak{p}}$$ is a domain. We want to show that the set $$D$$ of prime ideals $$\mathfrak{q}$$ of $$A$$ such that $$A_{\mathfrak{q}}$$ is an integral domain is open.

Let $$I \subset A$$ be the kernel of the localization $$A\rightarrow A_{\mathfrak{p}}$$, $$I$$ is generated by elements $$a_1,\ldots,a_n$$. For each $$i$$, there exists some $$b_i \notin \mathfrak{p}$$ such that $$a_ib_i=0$$. Let $$f=b_1\ldots b_n \notin \mathfrak{p}$$. Then the image of $$I$$ in $$A_f$$ vanishes.

Let $$x=a/b \in A_f$$ be such that its image in $$A_{\mathfrak{p}}$$ is zero. Then the image of $$a$$ in $$A_{\mathfrak{p}}$$ vanishes, so $$a \in I$$ and thus $$x=a/b=0$$.

In other words, the localization $$A_f \rightarrow A_{\mathfrak{p}}$$ is injective. As $$A_{\mathfrak{p}}$$ is an integral domain, so is $$A_f$$.

Now, if $$\mathfrak{q}$$ is another prime ideal of $$A$$ not containing $$f$$, then $$A_{\mathfrak{q}}$$ is a localization of the integral domain $$A_f$$ so is an integral domain. In other words, $$\mathfrak{p}\in D(f) \subset D$$. QED.

• Thank you! I didn't understand that $A_{\mathfrak{q}}$ is a localization of $A_f$ but now it's ok! Mar 28 at 13:38

I think this can be done with primary decomposition :)

Suppose $$X=\text{Spec}(A)$$ is an affine Noetherian scheme and set $$Z=\{\mathfrak{p}\in \text{Spec}(A)\mid A_\mathfrak{p}\text{ is not an integral domain}\}.$$ In general, all ideals $$\mathfrak{a}$$ in Noetherian rings admit some primary decomposition - $$\mathfrak{a}$$ can be expressed as an intersection of finitely many primary ideals $$\mathfrak{a}=\mathfrak{q}_1\cap\dots\cap\mathfrak{q}_n$$ whose radicals $$\mathfrak{p}_1=\sqrt{\mathfrak{q}_1},\dots,\mathfrak{p}_n=\sqrt{\mathfrak{q}}_n$$ are precisely the prime ideals which occur in the set $$\{\sqrt{(\mathfrak{a}:x)}\mid x\in A\}$$ where $$(\mathfrak{a}:x):=\{y\in A\mid yx\in\mathfrak{a}\}$$.

If you take a primary decomposition of the zero ideal in $$A$$ $$(0)=\mathfrak{q}_1\cap\dots\cap\mathfrak{q}_n$$ $$\mathfrak{p}_1=\sqrt{\mathfrak{q}_1},\dots,\mathfrak{p}_n=\sqrt{\mathfrak{q}_n}$$ we get that the set $$D$$ of zero divisors in $$A$$ $$D=\bigcup\limits_{x\in A\diagdown\{0\}}\,(0:x)$$ - which is evidently equal to its radical (an element is a zero divisor if an only if one of its powers is) $$D=\sqrt{\bigcup\limits_{x\in A\diagdown\{0\}}\,(0:x)}=\bigcup\limits_{x\in A\diagdown\{0\}}\,\sqrt{(0:x)}$$ - must be contained in the union $$\mathfrak{p}_1\cup\dots\cup\mathfrak{p}_n$$; conversely, each of the primes $$\mathfrak{p}_1,\dots,\mathfrak{p}_n$$ is made up of zero divisors since they're all of the form $$\sqrt{(0:x)}$$ for appropriate elements $$x\in A$$, whence $$D=\mathfrak{p}_1\cup\dots\cup\mathfrak{p}_n$$.

It follows that (since a primary decomposition of the zero-ideal in any localization $$A_\mathfrak{p}$$ is given by the intersection of the non-trivial extended primary ideals $$(\mathfrak{q}_1)_\mathfrak{p}\cap\dots\cap(\mathfrak{q}_1)_\mathfrak{p}$$) it follows that $$A_\mathfrak{p}$$ has no zero-divisors if and only each of the primes $$\mathfrak{p}_1,\dots,\mathfrak{p}_n$$ extends to the unit ideal in $$A_\mathfrak{p}$$, i.e. if and only if $$\mathfrak{p}$$ doesn't contain any of the primes $$\mathfrak{p}_1,\dots,\mathfrak{p}_n$$. In other words $$Z=V(\mathfrak{p}_1\cap\dots\cap\mathfrak{p}_n)$$.

Hope this helps and I didn't make any silly mistakes :p

• Thank you so much! I really like your solution. In some sense, it is close to the way I was following. Mar 28 at 12:01